$\overline{AB} = 5\sqrt{5}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $5\sqrt{5}$ $?$ $ \sin( \angle BAC ) = \frac{ \sqrt{5}}{5}, \cos( \angle BAC ) = \frac{2\sqrt{5} }{5}, \tan( \angle BAC ) = \dfrac{1}{2}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{5\sqrt{5}} $ $ \overline{AC}=5\sqrt{5} \cdot \cos( \angle BAC ) = 5\sqrt{5} \cdot \frac{2\sqrt{5} }{5} = 10$